Teenage alcoholism and weapon grade encephalitis.  Alcohol actually provides temporary relief.  In these cases, alcohol clears the mind.

Just Like A Snake

Temperature particles provide continuous spectra in the ultraviolet, the infrared, and visible range ("Lemmings Over The Cliff..." dated 18 May 2016).  Temperature particles could also be responsible for smell and taste, that fresh vegetable and food taste bland after they have been stored in cool temperature.   The coolness of negative temperature particles neutralizes  positive temperature particle and takes the taste/smell away.

So, if you are able to isolate temperature particles, give positive temperature particles a good lick and varies its amount as you do so.  It is likely that a combination of positive and negative temperature particles in succession provide a unique favor to foodstuff.

Just like a snake tasting the air with its tongue is seeing in infrared.

Absorbed!

But a plot of (x^(1/3)-a^(1/3))/((a*x)^(1/3))-ln((x/a)^(1/3)) for a =1 to 7 is,

All absorption lines!  A plot of 1/a^2-1/x^2 for a =1 to 7 gives,

which indicates,

$h\ne\cfrac{c}{5\pi }a_{\psi\,c}m_{ \psi  }$

on the basis that Rydberg formula is correct.

What is $h$?

It Is Just Nice To Look At

Hei, from the previous post "A Deep Dark Secret" dated 25 Oct 2016,

$F=\cfrac { 2 }{ 5 } m_{ \psi  }a^{ 2 }_{ \psi  }\left( \cfrac { c }{ 2\pi a_{ \psi  } }  \right) ^{ 2 }\cfrac{1}{a_{\psi}}$

instead,

$F_{\rho}=\cfrac { 2 }{ 5 } m_{ \psi  }a^{ 2 }_{ \psi  }\left( \cfrac { c }{ 2\pi a_{ \psi  } }  \right) ^{ 2 }\cfrac{1}{a_{\psi}}$

since we are dealing with $\psi$ and it is force density all along, $m_{\psi}=m$ is mass density, a constant.  So,

$F_{\rho\,n_1}=\cfrac { 2 }{ 5 } m_{ \psi  }a^{ 2 }_{ \psi \,n1 }\left( \cfrac { c }{ 2\pi a_{ \psi n1 } }  \right) ^{ 2 }\cfrac{1}{a_{\psi\,n1}}$

then

$F_{\rho\,n_1}=\cfrac{1}{10\pi^2 }\cfrac{m_{ \psi  }c^2}{a_{\psi\,n1}}$

and similarly,

$F_{\rho\,n_2}=\cfrac{1}{10\pi^2 }\cfrac{m_{ \psi  }c^2}{a_{\psi\,n2}}$

In general,

$F_{\rho\,n}=\cfrac{1}{10\pi^2 }\cfrac{m_{ \psi  }c^2}{a_{\psi\,c}}.\cfrac{1}{\sqrt[3]{n}}$

since,

$\cfrac{1}{n}=\left(\cfrac{a_{\psi\,c}}{a_{\psi\,n}}\right)^3$

as such,

$W_{1\rightarrow 2}=\int^{a_{ \psi\,n2 }}_{a_{ \psi\,n1 }}{F_{\rho\,n}}\,da_{\psi}$

$W_{1\rightarrow 2}=\int^{n2}_{n1}{F_{\rho\,n}}\,d\left(\sqrt[3]{n}.a_{\psi\,c}\right)$

$W_{1\rightarrow 2}=\cfrac{m_{ \psi  }c^2}{10\pi^2 }\int^{n2}_{n1}{\cfrac{1}{\sqrt[3]{n}}}.d\sqrt[3]{n}$

and

$W_{1\rightarrow 2}=\cfrac{m_{ \psi  }c^2}{10\pi^2 }\left[ln(\sqrt[3]{n_2})-ln(\sqrt[3]{n_1})\right]=\cfrac{m_{ \psi  }c^2}{10\pi^2 }ln\left(\cfrac{\sqrt[3]{n_2}}{\sqrt[3]{n_1}}\right)$

$W_{1\rightarrow 2}=\cfrac{m_{ \psi  }c^2}{10\pi^2 }ln\left(\cfrac{a_{\psi\,n2}}{a_{\psi\,n1}}\right)$

Which is somehow very satisfying, irrespective of all and any maths and logic blunders.  It is so nice I took a second look.  Since,

$f_c=\cfrac{c}{2\pi a_{\psi\,c}}$

the expression suggests,

$h=\cfrac{c}{5\pi }a_{\psi\,c}m_{ \psi  }$

which would be a constant for a given particle.  Nice!

A plot of ln((n_2/n_1)^(1/3)) for n_1 = 1 to 7 is given below,

As $a_{\psi\,n1}\rightarrow a_{\psi\,n2}$, there is an release of energy as the particle grows bigger.  $W_{1\rightarrow 2}$ requires energy and reduces the amount of energy released.

The most energy required from a transition from $n_1=1$ results in a absorption line.

Big particle with less energy in $\psi$ is counter-intuitive.  Big particle has $\psi$ going around its circumference at lower frequency, given that $\psi$ has a constant speed, light speed.

But a big particle contains a small particle!?

A big particle has more extensive $\psi$ but its $\psi$ has less energy when we take reference at the circumference/surface of the particle.

$v=r\omega$

All inner $\psi$ have lower speed, when they do not slide along each other.  $\psi$ outside from the center of the particle need to move faster; the fastest of which is light speed.

A Deep Dark Secret

Consider the force holding $\psi$ in circular motion,

$L=I\omega$

$F=L\cfrac{\omega}{r}$

scale by $r$, as the further from the center the less $L$ has to turn,

$F=I\cfrac{\omega ^{ 2 }}{r}$

In the case of a point mass, $m$ in circular motion with velocity $v$, in circle of radius of $r$,

$F=mr^2\left(\cfrac{v}{r}\right)^2.\cfrac{1}{r}=m\cfrac{v^2}{r}$

where $I=mr^2$ and  $\omega =\cfrac{v}{r}$ in radian per second.  Which is as expected.

In this case of a sphere of $\psi$,

$F=\cfrac { 2 }{ 5 } m_{ \psi  }a^{ 2 }_{ \psi  }\left( \cfrac { c }{ 2\pi a_{ \psi  } }  \right) ^{ 2 }.\cfrac{1}{a_{ \psi  } }$

where $\omega=\cfrac{v}{2\pi a_{\psi}}$ in per second.

$m_{ \psi  }=\rho _{ \psi  }.\cfrac { 4 }{ 3 } \pi a^{ 3 }_{ \psi  }$

$F=\cfrac { 4 }{ 30\pi }  c^{ 2 }\rho _{ \psi  }.a^{ 2 }_{ \psi  }$

$W_{ 1\rightarrow 2 }=\int _{ a_{ \psi \,n1 } }^{ a_{ \psi \, n2 } }{ F } da_{ \psi  }$

If we assume that, $\rho _{ \psi  }\propto \psi$.

For,

$a_{ \psi \, n1 }>a_{ \psi \,\pi  }$ and $a_{ \psi \, n2 }>a_{ \psi \,\pi  }$

$\psi=\psi_{\pi}=constant$

We have,

$W_{ 1\rightarrow 2 }=\cfrac { 4 }{ 90\pi } c^{ 2 }\rho _{ \psi  }\left[ a^{ 3 }_{ \psi \, n_{ 2 } }-a^{ 3 }_{ \psi \, n_{ 1 } } \right]$

$W_{ 1\rightarrow 2 }=\cfrac { 4 }{ 90\pi  } c^{ 2 }\rho _{ \psi  }a^{ 3 }_{ \psi \, c }\left[ n_{ 2 }-n_{ 1 } \right]$

which is just as wrong as the other expressions derived previously.

For,

$a_{ \psi \, \, n1 }\lt a_{ \psi \, \, \pi  }$

$\, \, a_{ \psi \, \, n2 }\lt a_{ \psi \, \, \pi  }$

$W_{ 1\rightarrow 2 }=\int _{ a_{ \psi \, \, n1 } }^{ a_{ \psi \, \, n2 } }{ F } da_{ \psi  }=\cfrac { 4 }{ 30 \pi} c^{ 2 }\int _{ a_{ \psi \, \, n1 } }^{ a_{ \psi \, \, n2 } }{ \rho _{ \psi  }.a^{2 }_{ \psi  } } da_{ \psi  }$

From the post "Not Quite The Same Newtonian Field" dated 23 Nov 2014,

$\psi =-i{ 2{ mc^{ 2 } } }\, ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_{ z })))+c$

$\psi =-{ 2{ mc^{ 2 } } }\, ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (a_{ \psi  })))$

under the assumption, $\rho _{ \psi  }\propto \psi$

$\rho _{ \psi  }=A. \psi$

We have,

$W_{ 1\rightarrow 2 }=A.\cfrac { 8 }{ 30 \pi } mc^{ 2 }\int _{ a_{ \psi \,  n1 } }^{ a_{ \psi \,  n2 } }{ -a^{ 2 }_{ \psi  } } ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (a_{ \psi  })))\,da_{ \psi  }$

At last, inevitably the ugly bride meets the in-laws ...

What is $m$?  This was a problem since $F_{\rho}$ or $F$, the force density was written down.  A force has to act on some mass.  If force density acts on mass density then,

$\rho_{\psi}=m$

then $\psi=f(\rho_{\psi})$, given $\rho_\psi$,

$\psi =-{ 2{ \rho_{\psi}c^{ 2 } } }\, ln(cosh(\cfrac { G }{ \sqrt { 2{ \rho_{\psi}c^{ 2 } } }  } (a_{ \psi  })))$

from which we solve for $\rho_{\psi}$ given $\psi$.  Which make sense because $W_{ 1\rightarrow 2 }$ is moving $\psi$ about, we must know the amount of $\psi$ in question to calculate $W_{ 1\rightarrow 2 }$.

But does energy density has mass density?  Does energy has mass?

This path does not provide an easy answer to $W_{ 1\rightarrow 2 }$  as $L_{n1}\rightarrow L_{n2}$.  $W_{ 1\rightarrow 2 }$ might be similar to Rydberg formula,

$\cfrac{1}{\lambda}=R_{\small{H}}\left(\cfrac{1}{n^2_1}-\cfrac{1}{n^2_2}\right)$

A new passport and a ticket to nowhere...you asked for it!

Not Orbits!

There is no orbits, $\psi$ is held by $F_{\rho}$ the force density.  Moving $\psi$ away from the center of the particle act against $F_{\rho}$ or $\int{F_{\rho}}dr$.

There is no charge particle holding another oppositely charged particle, and no field between the oppositely charged particles, along which potential energy changes.

Adjusting for Bohr - Angular Momentum

The expression,

$\Delta E_{2\rightarrow 1}=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1n_2}}-\cfrac { n_{ o2 }^{ 2 }-n_{ o1 }^{ 2 } }{ (n_{o1 }n_{ o2 })^{ 2 } }\right)$

from the post "Particles In Orbits" dated 18 Oct 2016, with $n_i=n_{oi}=i$,

$\Delta E_{2\rightarrow 1}=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1n_2}}-\cfrac { n_{ 2 }^{ 2 }-n_{ 1 }^{ 2 } }{ (n_{1 }n_{ 2 })^{ 2 } }\right)$

is ARBITRARY.  It is the result of comparing the leading constant to Planck relation $E=h.f$

The term,

$\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1n_2}}$

is due to the difference in potential energy associated with the spin of $\psi$ with $n_1$ and $n_2$, at $a_{\psi\,n_1}$ and $a_{\psi\,n_2}$, respectively, and the term,

$\cfrac { n_{ 2 }^{ 2 }-n_{ 1 }^{ 2 } }{ (n_{1 }n_{ 2 })^{ 2 } }$

is the change in energy as momentum changes due to a change in $n$, assuming that the particles after coalescence/separation are still at light speed.

Let's formulate the change in energy as per Bohr model due to a quantized change in momentum.  Since $n$ is an integer, the change in momentum is equally spaced and so if $L_{n1}$ is the momentum of the

$KE_{n1}=\cfrac{L_{n1}^2}{2m_1}$
$KE_{n2}=\cfrac{L_{n2}^2}{2m_2}$

the change in $KE$ is

$\Delta KE=KE_{n2}-KE_{n1}=-\Delta PE$

$\Delta PE=\cfrac{L_{n1}^2}{2m_1}-\cfrac{L_{n2}^2}{2m_2}$

since we know that the change in momentum is entirely due to a change in $n$.  For a particle spinning at light speed $c$,

$L_{ n2 }=\cfrac{2}{5}m_{ 2 }a^2_{ \psi \, \, n2 }.\cfrac{c}{2\pi a_{\psi\,\,n2}}=\cfrac{c}{5\pi}m_{ 2 }a_{ \psi \, \, n2 }$

$L_{ n1 }=\cfrac{2}{5}m_{ 1 }a^2_{ \psi \, \, n1 }.\cfrac{c}{2\pi a_{\psi\,\,n1}}=\cfrac{c}{5\pi}m_{ 1 }a_{ \psi \, \, n1 }$

where $I_{ni}=\cfrac{2}{5}m_i.a^2_{\psi\,\,ni}$ is the moment of inertia of a sphere of radius $a_{\psi\,\,ni}$.  So,

$\Delta PE=\cfrac { m_{ 1 }\left( a_{ \psi \, \, n1 }c \right) ^{ 2 } }{ 50 \pi^2} -\cfrac { m_{ 2 }\left( a_{ \psi \, \, n2 }c \right) ^{ 2 } }{ 50\pi^2 } =\cfrac { c^{ 2 } }{50\pi^2 } \left( m_{ 1 }a^{ 2 }_{ \psi \, \, n1 }-m_{ 2 }a^{ 2 }_{ \psi \, \, n2 } \right)$

As particle of radius $a_{ \psi \, \, n1 }$ is made up of $n_1$ basic particle of radius $a_{\psi\,c}$, $n=1$

$\cfrac { 1 }{ n_{ 1 } } =\left( \cfrac { a_{ \psi \, \, c } }{ a_{ \psi \, \, n1 } }  \right) ^{ 3 }$,   $a_{ \psi \, \, n2 }=\sqrt [ 3 ]{ n_{ 2 } } a_{ \psi \, \, c }$

and particle of radius $a_{ \psi \, \, n2 }$ is made up of $n_2$ basic particle of radius $a_{\psi\,c}$, $n=1$

$\cfrac { 1 }{ n_{ 2 } } =\left( \cfrac { a_{ \psi \, \, c } }{ a_{ \psi \, \, n2 } }  \right) ^{ 3 }$,   $a_{ \psi \, \, n1 }=\sqrt [ 3 ]{ n_{ 1 } } a_{ \psi \, \, c }$

Since we have assumed that the volume of the particles are conserved and their density is a constant,

$m_{ 1 }=n_{ 1 }m_{ c }$

$m_{ 2 }=n_{ 2 }m_{ c }$

So after substituting for $m_{i}$,

$\Delta PE=\cfrac { m_{ c }c^{ 2 } }{ 50\pi^2 } \left( n_{ 1 }a^{ 2 }_{ \psi \, \, n1 }-n_{ 2 }a^{ 2 }_{ \psi \, \, n2 } \right)$

And after substituting for $a_{\psi\,i}$,

$\Delta PE=\cfrac { m_{ c }c^{ 2 } }{50\pi^2 } \left( n_{ 1 }\left( \sqrt [ 3 ]{ n_{ 1 } } a_{ \psi \, \, c } \right) ^{ 2 }-n_{ 2 }\left( \sqrt [ 3 ]{ n_{ 2 } } a_{ \psi \, \, c } \right) ^{ 2 } \right)$

$\Delta PE=\cfrac { m_{ c }c^{ 2 } }{ 50\pi^2 } a^{ 2 }_{ \psi \, \, c }\left( n_{ 1 }^{ 5/3 }-n_{ 2 }^{ 5/3 } \right)$

$L_{ c }=\cfrac{2}{5}m_{ c }a^2_{ \psi \, c }.\cfrac{c}{2\pi a_{ \psi \, c }}$

$\Delta PE=\cfrac { L_c^{ 2 } }{ 2m_c } \left( n_{ 1 }^{ 5/3 }-n_{ 2 }^{ 5/3 } \right)$

What happened to $\cfrac{1}{n^2_1}-\cfrac{1}{n^2_2}$?  The above expression is the change in $KE$ required for the specified change in momentum, it is not the associated change in energy level of the particle in its field.

Consider again,

$L_{ n2 }=\cfrac{2}{5}m_{ 2 }a^2_{ \psi \, \, n2 }.\cfrac{c}{2\pi a_{\psi\,\,n2}}=\cfrac{c}{5\pi}m_{ 2 }a_{ \psi \, \, n2 }$

$\cfrac{L_{ n2 }}{L_{ n1 }}=\cfrac{m_2}{m_1}.\cfrac{a_{ \psi \, \, n2 }}{a_{ \psi \, \, n1 }}$

$\cfrac{L_{ n2 }}{L_{ n1 }}=\cfrac{n_2}{n_1}.\sqrt[3]{\cfrac{ n_2 }{ n_1 }}=\left(\cfrac{n_2}{n_1}\right)^{4/3}$

Which is not $\cfrac{L_{n1}}{L_{n2}}=\cfrac{n_1}{n_2}$ as in Bohr model of

$L=n\hbar$

And if we follow through the derivation for the energy difference between two energy levels, $n_1$ and $n_2$ we have,

$E_{\small{B}}=R_{\small{E}}\left(\cfrac{1}{\left(n_1^{4/3}\right)^2}-\cfrac{1}{\left(n_2^{4/3}\right)^2}\right)=R_{\small{E}}\left(\cfrac{1}{n_1^{2.667}}-\cfrac{1}{n_2^{2.667}}\right)$

this does not effect the first term, $\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1n_2}}=\cfrac{1}{\sqrt[3]{n_1}}-\cfrac{1}{\sqrt[3]{n_2}}$ associated with the decrease in potential energy of the system in circular motion, as $\psi$ move from $n_1\rightarrow n_2$.

$E_{\small{B}}$ derived above is the potential energy change in the field ($E\propto\frac{1}{r}$) that accompanies a change in the angular momentum of $\psi$ as the energy level transition $n_1\rightarrow n_2$ occurs.  $E_{\small{B}}$ is recovered from the amount of potential energy associated with the system in circular motion, released.

$C=\cfrac { \sqrt [ 3 ]{ n_{ 2 } } -\sqrt [ 3 ]{ n_{ 1 } }  }{ \sqrt [ 3 ]{ n_{ 1 }n_{ 2 } }  } -\cfrac { n_{ 2 }^{ 2.667 }-n_{ 1 }^{ 2.667 } }{ (n_{ 1 }n_{ 2 })^{ 2.667 } }$

Does a system in circular motion have potential energy $h.f$ in addition to the system $KE$ as quantified by its angular momentum?  The work done field against the force $-\frac{\partial\psi}{\partial r}$ as $\psi$ moves away from the center along a radial line, is strictly not $\propto\frac{1}{r^2}$ but, the Newtonian $F$,

$F\propto -\int{F_{\rho} dr}$

$F\propto -ln(cosh(r))$

$E\propto -\int{ln(cosh(r))} dr$

and things get very difficult.

A series plot of ((x)^(1/3)-(a)^(1/3))/(a*x)^(1/3)-((x)^(2.667)-(a)^(2.667))/(a*x)^2.667 for a =1 to 6 is given below,

where the plots for $n_1=2$ and $n_1=3$ is almost coincidental.

After adjusting for $2\rightarrow 2.667$, the plots are essentially the same except that the overlapping plots are not $n_1=2$ and $n_1=5$ but $n_1=2$ and $n_1=3$.

It seems that $n_1=2$ and $n_1=3$ are degenerate, instead.

The important points are: a pair of degenerate plots ($n_1=2$ and $n_1=3$) very close together occurs naturally, that the plot for $n_1=1$ is an absorption line and the rest of the plots $n_1\ge 2$ are the emission background.

Good night.

Note: The potential energy of a ball in circular motion held by a spring, is the energy stored in the spring as the spring stretches to provide for a centripetal force.  It is clear in this case, that the $KE$ of the ball, (equivalently stated as angular momentum) is a separate issue from the potential energy stored in the spring.  The energy required to increase the ball's $KE_1\rightarrow KE_2$ is not the same as the energy required to stretch the spring further as the ball accelerates from $KE_1\rightarrow KE_2$.

It could be that $\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1n_2}}$ defines the energy change needed at the perimeter, $a_{\psi\,\,n2}$ and $-\cfrac { n_{ 2 }^{ 2.667 }-n_{ 1 }^{ 2.667 } }{ (n_{ 1 }n_{ 2 })^{ 2.667 } }$ is the change in energy along a radial line from $n_1$ to $n_2$.

Note: Positive is emission line, and negative is absorption line

How Big To Be?

The emission spectrum is continuous with $n_2$ large.  As the total energy of the system increases with high electric field or high temperature, $n_1$ can take on smaller values.  The absorption lines with $n_1=1$ are distinctive against the background of continuous emission.  How small is $n_{1}$ to be?

$n_1=1$

In order to generate a continuous emission spectrum, how big is $n_2$ to be?

$n_2=n_{\small{large}}\gt\gt 77$

How to make $n_2$ big?  If $n_2$ is spinning about the center of the particle, will $n_2$ be big?  The centripetal force acting against a pinch force that pull $\psi$ away from the particle will allow more $\psi$ to attach itself to the particle and allows $n_2$ to increase.

Maybe...

Electrons May Not Be Involved!

With,

$C=\cfrac { \sqrt [ 3 ]{ n_{ 2 } } -\sqrt [ 3 ]{ n_{ 1 } }  }{ \sqrt [ 3 ]{ n_{ 1 }n_{ 2 } }  } -\cfrac { n_{ 2 }^{ 2 }-n_{ 1 }^{ 2 } }{ (n_{ 1 }n_{ 2 })^{ 2 } }$

Could it be that because the spectrum observations are conducted at high temperature,

$\cfrac { \sqrt [ 3 ]{ n_{ 2 } } -\sqrt [ 3 ]{ n_{ 1 } }  }{ \sqrt [ 3 ]{ n_{ 1 }n_{ 2 } }  }$

is due to temperature particles.  And,

$\cfrac { n_{ 2 }^{ 2 }-n_{ 1 }^{ 2 } }{ (n_{ 1 }n_{ 2 })^{ 2 } }$

orbiting particles not necessarily electrons.

$C=C_{\small{T}}+C_{\small{B}}$

where $C_{\small{T}}$ is due to the energy density $\psi$ of interacting temperature particles and $C_{\small{B}}$ due to Bohr model not necessarily of electrons, with quantized momenta.

Temperature particles orbiting around electron orbits that in resonance produce the infrared spectrum and the ultraviolet spectrum was proposed previously "Lemmings Over The Cliff..." dated 18 May 2016.  Spectra lines in the visible spectrum will come from transitions between energy levels in the ultraviolet spectrum.

Electrons may not be involved!  Unless the observations are also conducted in the presence of a high electric field, electron are not affected.  If this is the case, temperature should be raised without the use of an electric field.  Within the confines of the experiment, there should be only one strong field affecting one type of particles.

Let electrons not be involved.  Maybe then spectra lines can be indicative of the existence of temperature particles in orbit around electron orbits as proposed in the post "Capturing $T^{+}$ Particles" dated 15 May 2016.

In which case, electron ionization energies has no direct relation with spectra lines (due to temperature particles), not even in the simplest model of Hydrogen atoms.

Have a nice day...

When Wrong Is Right

I know...

Emission spectrum is taken to be the complement of the absorption spectrum.

If all energy transitions have its complement then, they should all cancel.  Dark lines should not be visible and the background emission spectrum against which they appears must also be accounted for.  Ambient light or indirect illumination alone does not generate the background emission spectrum.

Emission spectrum in this blog refers to the backdrop upon which the dark absorption lines appears.

Redefining everything to be right...When wrong is right, what's left?  Right??

Eliminate all indirect illumination, does fine gaps appears in the background emission spectrum?

Wrong does not make left right, it just leaves behind more questions...  More questions are what's left!

What "It"?

Given,

$C=\cfrac { \sqrt [ 3 ]{ n_{ 2 } } -\sqrt [ 3 ]{ n_{ 1 } }  }{ \sqrt [ 3 ]{ n_{ 1 }n_{ 2 } }  } -\cfrac { n_{ 2 }^{ 2 }-n_{ 1 }^{ 2 } }{ (n_{ 1 }n_{ 2 })^{ 2 } }$

when $n_2\lt n_1$, for absorption lines, $n_1\leftrightarrow n_2$,

$D=\cfrac { n_{ 2 }^{ 2 }-n_{ 1 }^{ 2 } }{ (n_{ 1 }n_{ 2 })^{ 2 } }-\cfrac { \sqrt [ 3 ]{ n_{ 2 } } -\sqrt [ 3 ]{ n_{ 1 } }  }{ \sqrt [ 3 ]{ n_{ 1 }n_{ 2 } }  } =-C$

And so we can expect the spectra line $n_{1\rightarrow 2}$ to cancel the reverse spectra line $n_{2\rightarrow 1}$.

For the sake of it....

The Total Picture

Why is emission that occurs at the same time as absorption lines almost continuous, whereas absorption lines are spaced far apart?

For each value of $n_2$, we have an absorption line.  All possible values of $n_2$ generates an absorption series.

With many $n_1=1$ plots corresponding to different allowable energy states, we have a number of absorption series that depict such plausible energy states.

Which is confusing because of the one added level of indirection.  For example,

$n_1=1$ for $n=1$

$n_1=1$ for $n=2$

and

$n_1=1$ for $n=3$

where the lowest energy level of a particle of two constituent basic particles is different (higher) than the lowest energy of a particle with three constituent basic particles.

What happened to $n_i=n_{oi}=i$?

This expression considers the lowest (first, $n_1=1$) energy level in a set of all particle sizes (all $n$).

Energy transitions occur for a particle given its number of constituent basic particle size $n$ with a particular first/lowest energy state.  The lowest energy level $n_1$, given $n$ is an absorption series given all values of  $n_2$.  Across all possible values of $n_1$ as particle $n$ size changes, we have different absorption series.

And the question was, why is it an absorption line?  Which leads to the question, are there fine gaps in the background emission spectrum?

Note:  In the plot above $n_1=1$ is indicative of the lowest energy level of a particular $n$.  Given all values of $n_2$, $n_1=1$ for a particular $n$ generates a absorption series.  When $n$ is higher $n_1=1$ is lower, with less energy.  Different $n$, generates different spectra series.

Shifty Spectra

Given equal probabilities that an emission or absorption occurs at all energy levels to all energy levels, why does not $-E_{1\rightarrow n}$, an emission line and $E_{n\rightarrow 1}$, where $n_2=n_1$, an absorption line, cancels?

When $n_2\lt n_1$, emission plots become absorption plots but the single absorption plot does not become an emission plot.  For the case of $n_1=1$ there cannot be a lower $n_2$, so there is no emission lines from $n_1$ to a lower energy level.  This does not mean that there is no emission lines  that would cancel the absorption lines from $n_1=1$.  A transition $E_{2\rightarrow 1}$ would generate a emission line that cancels the absorption line of $E_{1\rightarrow 2}$.

Do the forward transition and its the reversed transition cancels?  Only with numerical calculations can tell.

but we see that,

because the emission plots are not parallel for values of $n_2$ smaller than $n_2$ at the minimum point.  And in the following case its is not possible to tell graphically whether the forward transition and its reversed transition cancels,

Notice that $E_{2\rightarrow 3}$ is an emission line and $E_{3\rightarrow 2}$ is an absorption line.  All transitions to the left of the final energy state plot minimum point $n_{2\,\,min}$ have the opposite sign.  The corresponding reversed transition is at the minimum point and move downwards vertically to the final energy state.

It is also noted that for the plot $n_1=5$ another zero x-axis intercept occurs at $n_2=2$. And that for the plot $n_1=2$ another zero x-axis intercept occurs at $n_2=5$.

Both plots are coincidental.  This suggests that moving from $n_1=2$ to $n_2=5$ and in reverse, requires no net energy.  In moving from $n_1=$ to $n_2=5$, the loss in energy due to a decrease in $a_{\psi}$ at $n_2=5$ is made up for by the increase in orbital energy there, and vice versa.

It is possible that there is an energy gradient as the spectra line observations are being made.  When energy of the system is increasing, small particles tend to form and $n_1\gt n_2$.  When energy of the system is decreasing, big particles tend to form on separation after a collision and $n_1\lt n_2$.  In this way, $n_1\ne n_2$ and the emission lines are not coincidental with the absorption lines.  Emission and absorption that cancel do not occur with equal probability.

Which means, with no energy input to the system, the spectra line would disappear, but a different sets of lines reappears as the system cools.

Note: The difference between $\Delta E_{1\rightarrow 2}$ and $E_{1\rightarrow 2}$ is the overall change in energy state as demarcated by the definition of one single process and one energy state change of many in a more prolonged process.

In the case where separation follows coalescence,

$\Delta E_{1\rightarrow 2}\ne h.f$

instead there are both an emission line,

$E_{1\rightarrow n}$

and an absorption line,

$E_{n\rightarrow 2}$

As the small particle grows bigger its energy drops due to a larger $a_{\psi}$.

We All Have Our Say, Give And Take

We can do away with the notion of colliding atoms and think of the high energy experimental conditions as being conducive to the formation of small particles,

In the top most diagram, a small particle passes through a big particle with energy transitions $-E_{1\rightarrow n}$ and $E_{n\rightarrow 2}$ occurring, on entering the big particle and on leaving the big particle.  Both emission and absorption of energy occur in the same process.  The overall change in energy state is,

$\Delta E_{1\rightarrow 2}=E_{n\rightarrow 2}-E_{1\rightarrow n}$

but $-E_{1\rightarrow n}$ is energy emission and $E_{n\rightarrow 2}$ is energy absorption.

In the middle diagram, the small particle coalesce with the big particle and the lowest energy level attained is,

$n=n_{ large }+n_{ 1 }$

and the resulting bigger particle subsequently breaks into $n-n_2$ and $n_2$ particles where $n_2\ne n_1$.  This differ from the toppest case where $n\ne n_{ large }+n_{ 1 }$ but simply $n\gt n_1$.  The small particle passing through the big particle retains its distinctiveness because of its high momentum.  These two diagrams provide two scenarios as to what happened to the small particle inside the big particle.  In both cases a subsequent departure follows coalescence.

The third diagram shows a simple coalescence without a subsequent separation.  Only one singular energy emission occurs as $n_1\rightarrow n_{\small{large}}+1$.

All three process occur simultaneously.  Emissions form the colored background against which dark absorption lines show up in contrast.

Have a nice day.

Note:  This discussion is of all $n_1$, not just $n_1=1$.  The case of $n_1=1$ generates an absorption series given the two processes involved; changing $a_{\psi}$ and changing orbital energy level as according to Bohr model.

Speculating While Others Sleep...

Continued from the previous post "Particles In Orbits" dated 18 Oct 2016,

$n_1=n_{o1}=1$

where the lowest energy level coincides with the smallest particle.  In general,

$n_i=n_{oi}=i$

if we argue that given light speed a constant,  angular momentum at light speed increases proportionally with the number of constituent basic particle $n$.  This makes $n_{oi}$ from Bohr model coincidental with $n_{i}$, the number of constituent basic particles.

Which brings us to the emission plots $n_1=2$ and $n_1=5$,

When experimental conditions are such that $n_i\ne n_{oi}$, these two plots will split and emerge as two separate emission lines.  A new line mysteriously appears.

When an appropriate external field is applied the angular momentum of the particle changes in the direction of the field, $n_i\ne n_{oi}$.

This is not Zeeman effect nor Stark effect.

But an indication of the two processes involved as particles coalesce and break.

Speculating while others sleep...

The problem is $n_2=80$ is not large enough for the various plots to be on a plateau (approaching an asymptote).  Since, spectra lines are observed only in high energy conditions and we postulate that high energy conditions enables $n_2\rightarrow large$, then

$R_{\small{H}}=\cfrac{1}{2\pi a_{\psi\,c}}$

is still on the chopping block, where

$\Delta E_{2\rightarrow 1}=h.f_{\psi\,c}\left(\cfrac{1}{\sqrt[3]{n_1}}-\cfrac { 1 }{ (n_{o1 })^{ 2 } }\right)$

$\Delta E_{n\rightarrow 1}=h.f_{\psi\,c}\left(\cfrac{1}{\sqrt[3]{n_1}}-\cfrac { 1 }{ (n_{o1 })^{ 2 } }\right)$

and

$\Delta E_{n\rightarrow 2}=h.f_{\psi\,c}\left(\cfrac{1}{\sqrt[3]{n_2}}-\cfrac { 1 }{ (n_{o2 })^{ 2 } }\right)$

And in a sleep deprived and convoluted way,

$\Delta E_{2\rightarrow 1}-\Delta E_{n\rightarrow 2}=h.f_{\psi\,c}\left(\cfrac{1}{\sqrt[3]{n_2}}-\cfrac{1}{\sqrt[3]{n_1}}+\cfrac { 1 }{ (n_{o2 })^{ 2 } }-\cfrac { 1 }{ (n_{o1 })^{2 } }\right)$

Which is really interesting...a transit from energy level $1$ to a higher level $n$ and a return to energy level $2$.  The negative sign here makes the absorption line an emission line.  This is how an absorption spectrum has a parallel emission spectrum!

A collision not of basic particles but atoms that contains the basic particles.  The high energy level state is achieved when the colliding atoms are close together on impact.  As the atoms part after the collision, the transition $\Delta E_{n\rightarrow 2}$ occurs.

Hmmm...

Particles In Orbits

The difference plot of the expression, $\cfrac { \sqrt [ 3 ]{ n_{ 2 } } -\sqrt [ 3 ]{ n_{ 1 } }  }{ \sqrt [ 3 ]{ n_{ 1 }n_{ 2 } }  } -\cfrac { n_{ 2 }^{ 2 }-n_{ 1 }^{ 2 } }{ (n_{ 1 }n_{ 2 })^{ 2 } }$

however, draws a sense of déjà vu...

The concepts leading to the expression $\cfrac { \sqrt [ 3 ]{ n_{ 2 } } -\sqrt [ 3 ]{ n_{ 1 } }  }{ \sqrt [ 3 ]{ n_{ 1 }n_{ 2 } }  }$ reverses the energy sign of conventional electron energy level transitions; $E_{1\rightarrow 2}$ is negative and the particle loses energy.  $E_{1\rightarrow 2}$ is emitted

This would make the plot for $n_1=1$ in the above graph, an absorption line.

For all values of $n_2$, only when $n_1=1$ is an absorption line.  The plots in black are emission lines against which we see the absorption line without direct illuminations.

Values of the plots in black below $y=0$ is ignored because $n_2\ge n_1$.

Rydberg constant is not murdered, instead a new process is given birth.

$\cfrac { \sqrt [ 3 ]{ n_{ 2 } } -\sqrt [ 3 ]{ n_{ 1 } }  }{ \sqrt [ 3 ]{ n_{ 1 }n_{ 2 } }  }$ occurs at the same time as $\cfrac { n_{ 2 }^{ 2 }-n_{ 1 }^{ 2 } }{ (n_{ 1 }n_{ 2 })^{ 2 } }$ due to quantized energy levels in Bohr model theory.  The two process has reverse energy signs and hence married with a negative sign.  According to Bohr model $E_{1\rightarrow 2}$ is positive and the particle gains energy and transits to a higher energy level.

We have instead,

$\Delta E_{2\rightarrow 1}=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1n_2}}-\cfrac { n_{ o2 }^{ 2 }-n_{ o1 }^{ 2 } }{ (n_{o1 }n_{ o2 })^{ 2 } }\right)$

where a change $n_1\rightarrow n_2$ is accompanied by a change in orbital energy level $n_{o1}\rightarrow n_{o2}$.

Only now are the particles in orbits.

Note:  $E_2=E_1+\Delta E_{1\rightarrow 2}$

Murder Yet Written

Of course,

$\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1n_2}}\ne\cfrac{1}{n_1^2}-\cfrac{1}{n_2^2}=\cfrac{n_2^2-n_1^2}{(n_1n_2)^2}$

but are they parallel over the range of $n_1$ and $n_2$ in consideration.

A plot of $\cfrac { \sqrt [ 3 ]{ n_{ 2 } } -\sqrt [ 3 ]{ n_{ 1 } }  }{ \sqrt [ 3 ]{ n_{ 1 }n_{ 2 } }  } -\cfrac { n_{ 2 }^{ 2 }-n_{ 1 }^{ 2 } }{ (n_{ 1 }n_{ 2 })^{ 2 } }$ gives

where the values of the expression for $n_1=1$ is negative and the graphs of for $n_1=2$ and $n_1=5$ are coincidental.

A plot of the ratio $\cfrac { \cfrac { \sqrt [ 3 ]{ n_{ 2 } } -\sqrt [ 3 ]{ n_{ 1 } }  }{ \sqrt [ 3 ]{ n_{ 1 }n_{ 2 } }  }  }{ \cfrac { n_{ 2 }^{ 2 }-n_{ 1 }^{ 2 } }{ (n_{ 1 }n_{ 2 })^{ 2 } }  }$ gives,

The values of the ratio varies with both $n_2$ and $n_1$.  In the case of $n_1=1$, the ratio $\cfrac { \cfrac { \sqrt [ 3 ]{ n_{ 2 } } -\sqrt [ 3 ]{ n_{ 1 } }  }{ \sqrt [ 3 ]{ n_{ 1 }n_{ 2 } }  }  }{ \cfrac { n_{ 2 }^{ 2 }-n_{ 1 }^{ 2 } }{ (n_{ 1 }n_{ 2 })^{ 2 } }  }$ is close to $1$

Rydberg constant is not dead.  Yet...

Looking for Murder

From the previous post "" dated 13 Oct 2016,

$\Delta E_{1\rightarrow 2}=0.2063hf_{\psi\,c}$

and so we have,

$a_{\psi\,new}=0.2063a_{\psi}$

thus assuming that the first spectra line is of double intensity, ie from the pair $(n_2=2,n_1=1)$,

$a_\psi=0.2063*19.34=3.99\,nm$

$a_\psi=0.2063*16.32=3.37\,nm$

$a_\psi=0.2063*15.48=3.19\, nm$

$a_\psi=0.2063*14.77=3.05\,nm$

If, the first spectra line is the result of $(n_{\small{large}},1)$,

$a_\psi=0.762*19.34=14.74\,nm$

$a_\psi=0.762*16.32=12.44\,nm$

$a_\psi=0.762*15.48=11.80\, nm$

$a_\psi=0.762*14.77=11.25\,nm$

Which is just a bunch of numbers.  However, both cases point to the fact that the calculated $f_{\psi}$ from Planck's relation, $\Delta E=h.f_{\psi}$, is not the frequency $f_{\psi\,c}$ of $\psi$, to resonate the particle with.  $f_{\psi\,c}$ is defined as,

$f_{\psi\,c}=\cfrac{c}{\lambda_{\psi\,c}}=\cfrac{c}{2\pi a_{\psi\,c}}$

where there is one wavelength $m=1$ around the circular path of $\psi$ with radius $a_{\psi\,c}$ .

Depending on which spectra line is used to derive $f_{\psi}$, a factor of $0.2063$ or $0.762$ applies.  In the case where both the double intensity emission spectra line and the first absorption spectra line are used, with the appropriate factor, there can be only one value for $f_{\psi\,c}$.

Where do all these lead us?

$\Delta E_{2\rightarrow 1}=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1n_2}}\right)$

The murder of Rydberg constant?

$\cfrac{1}{\lambda_{2\rightarrow 1}}=R_{\small{H}}\left(\cfrac{1}{n_1^2}-\cfrac{1}{n_2^2}\right)$

Comparing the two expressions,

$R_{\small{H}}=\cfrac{1}{2\pi a_{\psi\,c}}$

which is a constant given $a_{\psi\,c}$.  Maybe...

Don't Worry, Be Creepy

It does not quite matter, given

$f(n)=\left(\cfrac{\sqrt[3]{n}-{1}}{\sqrt[3]{n}}\right)$

n	f(n)
70	0.7573572497
71	0.758501808
72	0.7596250716
73	0.7607277244
74	0.7618104192
75	0.7628737797
76	0.7639184021
77	0.7649448568
78	0.7659536896
79	0.7669454232

The value of $f(n)$ for the range $70\le n\le79$ average to

$f(n)=0.762$

from which we may estimate $f_{\psi\,c}$ from,

$\Delta E_{2\rightarrow 1}=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1}\sqrt[3]{n_2}}\right)$

where $n_1=1$

from the post "Sizing Them Up Again..." dated 4 Oct 2016.

But what is $n$ or $n_{\small{large}}$?

Better yet since the emission spectra line as the result of  $(n_2=2,\,n_1=1)$, (two basic particles, $n=1$ coalesce) that produces a pair of photons, $\Delta E_{1\rightarrow 2}$ has double the intensity and hence readily identifiable,

$\Delta E_{1\rightarrow 2}=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{2}-\sqrt[3]{1}}{\sqrt[3]{1}\sqrt[3]{2}}\right)$

where $n_1=1$ and $n_2=2$.

$\Delta E_{1\rightarrow 2}=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{2}-1}{\sqrt[3]{2}}\right)$

$\Delta E_{1\rightarrow 2}=0.2063hf_{\psi\,c}$

without worry about what $n_{\small{large}}$ might be.

This is different from Planck's relation $E=h.f$.  A factor of about $\frac{1}{5}$ creeps in.

The First Spectra Line

For the absorption spectrum where energy is measured with reference from a higher threshold, the highest energy level has the lowest drop from such a reference, we reflect the previous plot from the post "Emission Spectrum Simplified" dated 12 Oct 2016, about the axis $y=0$.

The first spectra line is not the line with double intensity, but

$\Delta E_{1\rightarrow n=large}=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{n_{\small{large}}}-\sqrt[3]{1}}{\sqrt[3]{1}\sqrt[3]{n_{\small{large}}}}\right)=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{n_{\small{large}}}-1}{\sqrt[3]{n_{\small{large}}}}\right)$ --- (*)

To see what is $\left(\cfrac{\sqrt[3]{n_{large}}-{1}}{\sqrt[3]{n_{large}}}\right)$, we plot

As $n_{\small{large}}\rightarrow 77$, the increment in$\left(\cfrac{\sqrt[3]{n_{large}}-{1}}{\sqrt[3]{n_{large}}}\right)$ decreases but the plot is not asymptotic towards a steady value around $n=70\approx 80$.  (When $n\rightarrow \infty$, we have an asymptote towards $y=1$.)

Given that $n$ takes on integer values ($n+1$ being the number of constituent basic particles the big particle has before its breakage into a basic particle ($n=1$) and another big particle, $n$), and that expression (*) at high consecutive values of $n$ has very close values, spectra lines involving high values of $n$ will seem to split into numerous close lines.

This could be the explanation for split spectra lines.  The above is an emission spectra line plot, NOT an absorption spectra line plot.

To be sure that the first spectra line is given by (*), we look at other coalescence/breakage possibilities.  When big particles of $n+1$, $n+2$ and $n+3$ constituent basic particles break into pairs of $(n,\,1)$, $(n,\,2)$ and $(n,\,3)$  particles respectively,

Since the particle pairs $(n=3,\,n+3)$ are at closer energy levels than $(n=1,\,n+1)$, the transition from $n+3$ to $n=3$ requires less energy than the transition fro $n+1$ to $n=1$.  As such breakage into a small particle of higher $n$ involve lower absorption energy.  Such lines will be higher up in the absorption spectrum plot when this plot is reflected about $y=0$.

After considering other possible particle pairs, $(n=i,\,n+i)$, the first spectra line is still given by (*).

But how large is $n_{\small{large}}$? $n_{\small{large}}=77$? $n_{\small{large}}=74$?...

Emission Spectrum Simplified

Unfortunately in this scheme of things, we have $\psi$ balls of various sizes, $n$ in collisions.

When 2 particles of size $n=1$ coalesce 2 photons $E_{1\rightarrow2}$ are emitted, and when 2 particles of sizes $n=1$ and $n=2$ coalesce 2 different photons $E_{1\rightarrow 3}$ and $E_{2\rightarrow 3}$ are emitted.

This is not the simple scheme where emissions as the result of hops from energy level to energy level has equal intensity, but in the example above, $E_{1\rightarrow 2}$ has twice the intensity of $E_{1\rightarrow 3}$ and $E_{2\rightarrow 3}$.

The plot below shows ((x+1)^(1/3)-x^(1/3))/(((x+1)*x)^1/3), ((x+2)^(1/3)-x^(1/3))/(((x+2)*x)^1/3) and ((x+3)^(1/3)-x^(1/3))/(((x+3)*x)^1/3).

where particle of size $n=1$, $n=2$ and $n=3$ coalesce with particle of size $n=x$.

We see that the highest energy transition occurs with $E_{1\rightarrow n=large}$, when a basic particle $a_{\psi\,c}$ (ie. $n=1$) coalesces with a large particle $n\rightarrow 77$.  Two photons are released $E_{1\rightarrow n=large}$ and $E_{large\rightarrow large+1}$.

Emission occurs in bands as $n=x$ increases and such bands narrows with increasing $n=x$.

In the graph above, the top most three horizontal lines maroon, red and blue correspond to $x=1$.  The next band of maroon, red and blue correspond to $x=2$.  Each band progressively narrows as $x$ increases.

As $x$ increases, all graph approaches asymptotically to zero, ie as $x$ increases all emissions due to the coalescence of particles of various sizes, $n$ approaches zero.

Good night.

Sizing Them Up Again...

With this new picture of particles coalescence and disintegration, we will calculate $a_{\psi\,n=1}$ again...

From the post "No Nucleus Needed" dated 03 Oct 2016,

$\Delta E_{2\rightarrow 1}=E_{n_1}-E_{n_2}=h\cfrac{c}{2\pi a_{\psi\,n_1}}-h\cfrac{c}{2\pi a_{\psi\,n_2}}$

where positive values for $\Delta E_{n}$ is energy absorbed.  And from the post "Speculating About Spectra Series" dated 29 Sep 2016,

$a_{\psi\,n}=\sqrt[3]{n}.a_{\psi\,c}$ --- (*)

where  $n=1,\,2,\,3,..77$,

we have,

$\Delta E_{2\rightarrow 1}=h\cfrac{c}{2\pi }\left(\cfrac{1}{a_{\psi\,n_1}}-\cfrac{1}{a_{\psi\,n_2}}\right)=\hbar.c\left(\cfrac{a_{\psi\,n_2}-a_{\psi\,n_1}}{a_{\psi\,n_1}.a_{\psi\,n_2}}\right)$

Substitute (*) in,

$\Delta E_{2\rightarrow 1}=\hbar\cfrac{c}{a_{\psi\,c}}\left(\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1}\sqrt[3]{n_2}}\right)$

since,

$2\pi a_{\psi\,c}=\lambda_{\psi\,c}$  with $m=1$

assuming that there is $m=1$ wavelength along the circular path,

$\Delta E_{2\rightarrow 1}=h.f_{\psi\,c}\left(\cfrac{\sqrt[3]{n_2}-\sqrt[3]{n_1}}{\sqrt[3]{n_1}\sqrt[3]{n_2}}\right)$

where  $f_{\psi\,c}=\cfrac{c}{\lambda_{\psi\,c}}$.

Each possible value of $\Delta E_{2\rightarrow 1}$ due to a pair $\left(n_2,\,\,n_1\right)$ of particles before and after a collision, corresponds to one spectra line.  As more possible tuples of $\left(n_2,\,\,n_1\right)$ are made available given the set of experimental conditions, more spectra lines appear.

Now for experimental data from the web...

Where Small is Highest

The smaller the particle the higher $E_n$ is.  Big particle absorb energy to break into smaller particles.  When small particles coalesce energy is released as a photon that we observe as the emission spectrum.

Big particles are more susceptible to collision impacts and break into smaller particles, absorbing part of the energy of impact on breaking up.  Energy is absorbed as a photon, from which we obtain the absorption spectrum.

But $E_n$ is not $\psi$.  $E_n$ is the result of $\psi$ on a circular path at light speed $c$.

So,

$\lambda_\psi\ne\lambda_n$

and the photon packets emitted or absorbed,

$h.f_{vis}=h.\left(f_{n_1}-f_{n_2}\right)$

$\lambda_{vis}=\cfrac{c}{f_{vis}}=\cfrac{c}{f_{n_1}-f_{n_2}}$

where $f_{vis}$ and $\lambda_{vis}$ are obtained from the experimental spectrum(s) observed.

Obviously,

$\lambda_{vis}\ne\lambda_n$  and

$\lambda_{vis}\ne \lambda_{\psi}$

But,

$\lambda_{\psi\,n}=\lambda_{n}$,  $m=1$

only when $m=1$ that there is one wavelength around the circular path of radius $a_{\psi}$.  The factor $2\pi$ appears as the wavelength, $\lambda_{\psi}$ is wrapped around a circular path.

And since the energy transitions as particles coalesce and disintegrate is the reverse of electron energy level transitions, we do not have an issue with negative energy.

And in the rest state of no collisions, we have $a_{\psi\,n=1}$ where $a_{\psi}$ is the smallest at the highest energy possible,

$E_{n=1}=E_{max}$

$a_{\psi}=a_{\psi\,1}=a_{\psi\,c}$

So, paradoxically $\psi$ with $a_{\psi\,c}$ has the highest energy but the smallest size.

No Nucleus Needed

$\psi$ wraps around a sphere.  A certain amount of energy is associated with $\psi$ motion around the sphere.  If $h$ amount of energy is in the system, going once per second around the sphere, then travelling at light speed $c$ around a radius of $a_{\psi\,n}$, the amount of potential energy associated with this motion is,

$E_n=h.\cfrac{c}{2\pi a_{\psi\,n}}$

but,

$2\pi a_{\psi\,n}=m\lambda_m$

where $m$ is the number of wavelength, $\lambda_m$ of $\psi$ around the circle.

With $m=1$,

$E_n=h.\cfrac{c}{\lambda_m}=h.f$

which is Planck's relation given $m=1$.  When $\psi$ is a half wave in time and a half wave in space space,

$m=\cfrac{1}{2}$

$E_n=h.\cfrac{2c}{\lambda_m}=2h.f=h^{'}.f$

with a mysterious factor of $2$.

When $m=2, \,3...$

$E_n=h.\cfrac{c}{m\lambda_m}=\cfrac{h}{m}.f=h_m.f$

$h_m$ changes with $m$ because $c=f\lambda$.

And,

$h_{m_1}.f\gt h_{m_2}.f$

when $m_2\gt m_1$.

When there is more wavelengths around the circular path, stored energy decreases, which suggest another way $\psi$ can absorb and release packets of energy; by changing $m$.  Also, given $a_{\psi\,n}$ at fixed value, $f$ increases with $m$ but $h_m$ decreases, so, $E_n=h_m.f$ remains a constant.  $f$ increases with decreasing $a_{\psi}$, with increasing $E_n$.  Catastrophe when the number of particles with small $a_{\psi}$ is also large.

$E_n$ is independent of $m$.  $h$ is defined as one wavelength going around once per second along the circular path, then a corresponding increase in the number of wavelengths, $m$ increases the frequency by $m$ but this increase cancels with the decrease in $h$ by the reciprocal of $m$.

With these issues with $m$ aside, when two particles coalesce,

$a_{\psi\,n_1}\rightarrow a_{\psi\,n_2}$ 

the bigger particle has less potential energy associated with its $\psi$ in circular motion,  The amount of energy released due to this drop in potential is.

$\Delta E=E_{n_1}-E_{n_2}=h\cfrac{c}{2\pi a_{\psi\,n_1}}-h\cfrac{c}{2\pi a_{\psi\,n_2}}$

$\Delta E=hf_{n_1}-hf_{n_2}=h\left(f_{n_1}-f_{n_2}\right)$

$f_{n_1}\gt f_{n_2}$

using $h$ to denote all cases of $m$, as discussed above, we have an expression similar to the discrete energy level transitions expression for electrons around a nucleus,  In this case, however it is just $\psi$ alone going around a circular path and energy level decreases with increasing $a_{\psi\,n}$

No nucleus needed.

Spectra Ghost At The Rear

Happy Birthday to me...03 Oct 1968.

If spectra lines reflect the collision and subsequent coalescence of particles, then the observation that more spectra lines appear at lower experiment setup temperature, can be explained by lower collisions rate at lower temperature with less momentum which allows bigger particles to form.  Bigger particles have higher $a_{\psi\,n}$ which appear as a higher spectra line.  At higher temperature, big particles are broken up in high velocity impacts before further coalescence forms bigger particle.

So, at low temperature more spectra lines due to bigger particles appear.

Have a nice day.

The Quantum

From the previous post "Speculating About Spectra Series" dated 29 Sep 16,

$a_{\psi\,n}=\sqrt[3]{n}.a_{\psi\,c}$

which suggests energy transition,

$a_{\psi\,n_1}\rightarrow a_{\psi\,n_2}$

$n_1\rightarrow n_2$

occurs as small particles coalesce into bigger particles and when a big particle breaks into smaller particles.

The second case explains the occurrence of two or more simultaneous energy transitions, as a bigger particle breaks into two or more smaller particles.

$a_{ \psi \, 6 }\begin{matrix} \nearrow  \\ \rightarrow  \\ \searrow  \end{matrix}\begin{matrix} a_{ \psi \, 1 } \\  \\ a_{ \psi \, 2 } \\  \\ a_{ \psi \, 3 } \end{matrix}$

$n_{ 6 }\begin{matrix} \nearrow  \\ \rightarrow  \\ \searrow  \end{matrix}\begin{matrix} n_{ 1 } \\  \\ n_{ 2 } \\  \\ n_{ 3 } \end{matrix}$

In the case above, a particle made up of $n=6$ basic particles breaks into particles of size $n=1$, $n=2$ and $n=3$.  $n$ denoting arbitrary energy levels before has now a physical interpretation; it is the number of constituent basic particles.

I have found the quantum!  $a_{\psi\,c}$ is the quantum.